The radii of a frustum of a cone are 9, cm an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Radii $r_1 = 9\, cm$,$r_2 = 3\, cm$,and height $h = 8\, cm$.First,calculate the slant height $l$ using the formula $l = \sqrt{h^2 + (r_1 -

Vedclass · Accepted Answer

$376.8\, cm^2, 979.68\, cm^3$

Vedclass · Answer

(A) Given: Radii $r_1 = 9\, cm$,$r_2 = 3\, cm$,and height $h = 8\, cm$.First,calculate the slant height $l$ using the formula $l = \sqrt{h^2 + (r_1 - r_2)^2}$.$l = \sqrt{8^2 + (9 - 3)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\, cm$.Curved Surface Area $(CSA)$ of the frustum is given by $\pi(r_1 + r_2)l$.$CSA = 3.14 	imes (9 + 3) 	imes 10 = 3.14 	imes 12 	imes 10 = 376.8\, cm^2$.Volume of the frustum is given by $\frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1r_2)$.$V = \frac{1}{3} 	imes 3.14 	imes 8 	imes (9^2 + 3^2 + 9 	imes 3) = \frac{1}{3} 	imes 3.14 	imes 8 	imes (81 + 9 + 27) = \frac{1}{3} 	imes 3.14 	imes 8 	imes 117$.$V = 3.14 	imes 8 	imes 39 = 979.68\, cm^3$.

The radii of a frustum of a cone are $9\, cm$ and $3\, cm$. If its height is $8\, cm$,find its curved surface area and volume. $(\pi=3.14)$

Similar Questions

Total surface area of a closed cube with length $4 \, cm$ is $\ldots \ldots \ldots cm^{2}$.

The volume of a cone with radius $5 \, cm$ and height $12 \, cm$ is ........ $\pi \, cm^{3}$.

The volume of a hemisphere is equal to:

The volume of a sphere with radius $1.5\,cm$ is $\ldots \ldots \ldots \ldots cm^{3}$. (in $\pi$)

The radius and height of a cone are both $3 \, cm$. Then,its volume is $\ldots \ldots \ldots \, cm^{3}$. (in $\pi$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module